groups-rings-fields

[dull ginkgo]

it is a monoid though

[delicate orchid]

is your book looking at group actions from a categorical perspective or something

[dull ginkgo]

Mathematical crippling

[wraith swan]

It's explaining group theory with category theory

[delicate orchid]

is this your introduction to group theory or are you doing this just for funsies - also what's the name of the book? I'm curious

[wraith swan]

Chapter 0, Aluffi

Yeah it's my intro to group theory. And just for fun

[delicate orchid]

fair enough

[wraith swan]

I read into some HoTT for fun before, but it's a little abstract without any other mathematical knowledge

[dull ginkgo]

Aluffi is a funny name

[wraith swan]

I did read about category theory before but that had mostly examples in Set... Well and Cat but well, without concrete examples it's not easy to understand. Some people said to look at Abstract Algebra first

[coral spindle]

I hope you get something out of it, but please don't write off the theory as a whole before you've seen the non-categorical intro!

[wraith swan]

Oh I think it's pretty fun. Already gave some more insights into CT and even HoTT

[wraith swan]

Well it's not just a categorical intro, it's from both views I think

[hidden wind]

beware aluffi’s approach is very nonstandard

[wraith swan]

Next page he describes it as an operation on sets, right?

[dull ginkgo]

BE NOT AFRAID ass shit

[wraith swan]

But is non standard a problem?

Since I read something about CT in think it is fun to explore both at the same time

[agile burrow]

I like aluffi

[wraith swan]

Instead of reading about AA and learning CT Awodey style...

[agile burrow]

That's the book I learned algebra from

[wraith swan]

Cool

[crystal turtle]

he defines stuff using the most important definition (categories) so i like it

sotrue!

[winged void]

I have question is this enough

Let G act transtively on X, and let N be a normal subgroup of G. Prove that the orbits of X under N are all the same size; in other words: for all x and y in X, $#Nx$ = $#Ny$.
\begin{proof}
First of all, we know that N is a subgroup of G and it is normal, so this means for all $g\in G$ and $h \in N$ the following applies $ghg^{-1} \in N$. We know that $G$ is transitive, so this means that there is one orbit. Take two elements from $N_x$, for example $x$ and $x_1$, then the following applies
\begin{equation*}
h \circ x = x_1 \text{ for a $h \in N$ }
\end{equation*}
We also know that $G$ is transitive, so this means that
\begin{equation*}
g \circ x = y \text{ for a given $g \in G$}
\end{equation*}
We now know that the following must apply:
\begin{equation*}
x = g^{-1} y
\end{equation*}
We can now substitute this above and we get the following:
\begin{equation*}
h \circ g^{-1} y = x_1 \leftrightarrow h g^{-1} \circ y = x_1
\end{equation*}
It follows from this:
\begin{equation*}
g h g^{-1} \circ y = g\circ x_1
\end{equation*}
note that $ghg^{-1} \in N$ so this now means that:
\begin{equation*}
n\circ y = g\circ x_1
\end{equation*}
\end{proof}

[cloud walrus]

Mootje

[rocky cloak]

Looks good, but it would be more readable if you said a little bit more about what you're doing.

Like presumably you're saying that multiplication by g defines a bijection between Nx and Ny, so maybe at least say that

Then it can also be much shorter, like gNx = (gN)x = (Ng)x = N(gx) = Ny

[winged void]

But does this implies then they are equal in size

But does this implies then they are equal in size

That’s what I’m afraid about

[rocky cloak]

If you have a bijection between two sets, they are equal in size yes.

Often that's actually the definition of two sets having the same size

[round hull]

i have this elementary question

[dull ginkgo]

quasi semi group

[round hull]

let x, y be nonzero divisors of R. then (R/xR)/(yR/xR) is just R/(x, y)R, right?

[dull ginkgo]

Commutative ring?

[round hull]

this should be a statement about the fact that "quotients commute"

ya

but i don't see it exactly

[dull ginkgo]

That’s third iso I think

[delicate orchid]

feel that? that's the third iso theorem

[round hull]

sure something like that

[dull ginkgo]

Sniped

[round hull]

i want the fancy way though

[dull ginkgo]

(x,y) = (x) + (y) no?

[round hull]

ya

[dull ginkgo]

what happens when you quotient out (x) or (y) then

[round hull]

that also works

or i mean that is what you mean by third iso

[delicate orchid]

Doesn't third iso give us (R/xR)/(yR/xR) \cong R/yR though

[dull ginkgo]

It does ye

[delicate orchid]

exactly by this logic

ok so you're not saying it is R/(x,y)R

[round hull]

what is this cursedness

xR is not contained in yR

[delicate orchid]

then how are you quotienting?

your original question isn't well defined then

[round hull]

R/xR is an R-module

so you take y(R/xR)

[dull ginkgo]

[delicate orchid]

[dull ginkgo]

wew edit that gif of the guy releasing the trunk monkeys to be sully emojis

[delicate orchid]

anyway I realise that, if it was a quotient of rings, that this would just be yR because xR \subseteq yR

so it makes sense there

[round hull]

what's wrong with what i said

since R/xR is an R-module y gives an action on R/xR

and y(R/xR) is defined to be the image of that action

which is a submodule of R/xR

which we quotient by

[delicate orchid]

it was absolutely not clear at all that this was a quotient of modules

you could've just been working with non-unital rings

which is what we both assumed was happening

[round hull]

is there another way where yR/xR makes sense?

i see

[delicate orchid]

wait... even for modules you can't quotient by something that isn't a submodule

[round hull]

is there a non submodule somewhere

[winged void]

sure thanks alot

I have one more question

im suppose to prove that a 4d dimensional cube has 192 symmetris

or with other words that he has 8 elemnts of orbit and 24 elements of stabilizator

is there an easy way to do that

[delicate orchid]

192? that seems strange.

[rocky cloak]

Well, the symmetry group of a tesseract act transitively on the faces, and there are 8 faces. Each face is a cube and they have 24 symmetries

[delicate orchid]

jagr am I skill issuing here? It's definitely S_2 \wr S_4 right

[winged void]

So does that mean that if a cube has 24 symmetry then the 4d cube has also 24

[delicate orchid]

jagr practically told you that it's 8*24, which is 192

24 permutations of each face, and then 8 faces for a total of 192

[winged void]

i know that

but how to prove that

which approach should i follow

[rocky cloak]

With reflections maybe?

[delicate orchid]

yeah that's with reflections

ok if cubes have 24 symmetries we're definitely without reflections

mb

[delicate orchid]

do you know that the group of rotations of a cube is S_4?

[winged void]

Sure I agree

But I’m suppose to prove that as well

[delicate orchid]

the canonical way to do that is to look at how the rotations act on the 4 main diagonals of the cube

but there might be an easier way

yeah there is, I think

[rocky cloak]

If you just need that there's 24 of them, then you can repeat the argument I guess

[delicate orchid]

yeah, but then do we need that there are four rotations of a square?

I'm gonna exposit the way I just came up with because I think it's nice

nvm it's basically just the "permute the diagonals" argument in fancy clothes

[winged void]

Well then I do not need to use S4

Or I cannot I mean

[cobalt coral]

hi guys sorry to interrupt, needed some help with a concept. Im doing quotient groups of rings and have always been given distinct equivalence classes for quotient groups by my professor. Im not sure how to get these, or how to tell how many equiv classes there should be for a given quotient group

any help would be super appreciated 😦

[delicate orchid]

equivalence classes for quotient groups are exactly the cosets

you're quotienting by the relation g ~ g' iff gH = g'H, so the equivalence class of g is gH

[cobalt coral]

this is the problem i dont really understand

for polynomial rings do i divide the polynomials and find the euiv classes that way?

[delicate orchid]

who the hell is a .HEIC

[cobalt coral]

😭

[delicate orchid]

u running discord on a CDC 7600 or something

[delicate orchid]

no, the equivalence classes are the sets a+(x^2+x+1)

you can't divide polynomials in Z_2[x], it isn't a field

[cobalt coral]

ohhhhhh

thank u sooo much!!!!

[south patrol]

Source

[delicate orchid]

I'll do this REALLY explicitly if you want

[cobalt coral]

no, its all right

thank you!

[delicate orchid]

the idea behind quotients is that you're "adding relations in" to your object

so by quotienting out by (x^2+x+1), you're essentially saying "give me Z_2[x], but set x^2+x+1 = 0"

which is why all of our equivalence classes are only linear polynomials! because as soon as you see an x^2 you can go "hmm... since x^2+x+1 = 0, x^2 = x+1 (remember we're in Z_2 here, so -x = x)" and replace x^2 with a linear polynomial

[cobalt coral]

thank u so much ur a life saver

[delicate orchid]

https://ncatlab.org/nlab/show/field

[south patrol]

Finitely accessible

[round hull]

lol

"playing pretend"

good mo post

[delicate orchid]

it's the correct way to think about quotients

they turn equivalence relations into actual equality

[hidden wind]

it is something one should be aware of, and of course one cannot proclaim one approach better another without knowing both

[hidden wind]

i’ve been reading coxeter and moser’s generators and relations and struggling with making sense of this

esp with like making sense of what exactly is the relationship between the hyperbolic triangle reflection groups Δ(infty,infty,infty), Δ(2,3,infty) and SL(2,Z)

i just know the modular group is isomorphic to a index two subgroup of Δ(2,3,infty), namely the rotational or von Dyck triangle group D(2,3,infty)

which is wack

[delicate orchid]

oh lord

I don't know any hyperbolic geometry so I can't help you, sorry

[hidden wind]

me neither, i feel

[delicate orchid]

oh wait hold on

no I'm thinking of affine reflection groups

which is the most boring case of what you're talking about

[hidden wind]

well

[agile burrow]

I thought (2, 3 infty) is the modular group

[hidden wind]

it is

[agile burrow]

but index 2 here?

[hidden wind]

yes

[agile burrow]

😮

[hidden wind]

some people are not so explicit whether they are talking about the von Dyck or the full triangle group

[agile burrow]

oh maybe idk what triangle is vs von dyck

[delicate orchid]

I'm always suspicious of SL(2,Z). It's an omen for weird things

[hidden wind]

that would probably be Δ(infty,infty,infty) which is the free group over three order two generators

[wraith swan]

ok, but the math doesn't change, right? 😛

[delicate orchid]

free abelian surely?

[agile burrow]

ohh so von dyck subgroup is the orientation preserving stuff

[hidden wind]

no

[delicate orchid]

or am I visualising this wrong

[hidden wind]

yes

[delicate orchid]

ahhhh no you're right, there's a "sign error". And by that I mean the orientation is different

[hidden wind]

dw i prolly am too

[hidden wind]

typically these are presented purely abstractly anyways

[delicate orchid]

that's boring! I like picturing one of those scary fair ground mirror mazes

Δ(infty,infty,infty) would be like being stuck in a cube of mirrors... most terrifying indeed...

[hidden wind]

Δ(l,m,n) is < L, M, N | L^2 = M^2 = N^2 = (LM)^l = (MN)^m = (NL)^n = 1 >

[delicate orchid]

oh they're just coexter groups

[hidden wind]

aye

[delicate orchid]

duh. of course they are. They're reflection groups

[hidden wind]

and x^infty = 1 is vacous, so no relations on x

[delicate orchid]

yur

[hidden wind]

so Δ(infty,infty,infty) is < L,M,N | L^2 = M^2 = N^2 = 1 >

[delicate orchid]

Δ(3,3,3) is S_4 right?

[sly crescent]

IIRC this is PGL(2,Z)

[delicate orchid]

this is jolly good fun...

[sly crescent]

No

[delicate orchid]

WHAT

[sly crescent]

That’s Δ(2,3,3)

[delicate orchid]

rats

[hidden wind]

that’s a euclidean reflection group… which i’ve honestly spent way too little time on

[delicate orchid]

oh yeah duh. (12) and (34)

[hidden wind]

Δ(2,3,5) is also a nice one

[delicate orchid]

I swear I read the wikipedia page about these like years ago

[hidden wind]

may be viz like the second of these: #math-discussion message

[sly crescent]

The finite triangle groups are precisely (2,2,n), (2,3,3), (2,3,4), and (2,3,5)

[delicate orchid]

are these the ones for positive curvature space

well they'd have to be right?

[sly crescent]

Yes

[delicate orchid]

or else you just keep going foreverrrr

[hidden wind]

they provide tesselations of the sphere ye

[sly crescent]

The are the symmetry groups of the dihedra and hosohedra, the tetrahedron, the cube and octahedron, and the dodecahedron and icosahedron, respectively

[hidden wind]

this thing has Δ(2,3,5) symmetry

[delicate orchid]

I need to learn about these groups. This is so cool

[sly crescent]

This is wrong btw

[delicate orchid]

something something central extenstion of PSL(2,Z)

something something C_3 \ast C_2

[hidden wind]

me trying to make sense of these withiut knowing elementary group theory

[delicate orchid]

do you study combinatorics or something

[hidden wind]

oh absolutely not

[delicate orchid]

shame, coexter groups are very nice combinatorially

[sly crescent]

(2,3,∞) is PGL(2,Z)
(∞,∞,∞) has the free group on two generators as an index-2 subgroup, but I don’t remember whether or not it’s the direct product

[delicate orchid]

I wonder what happens if you... put more numbers in...

obviously it's n-dimensional space at that point but

[sly crescent]

Then you get Coxeter groups of higher rank

[delicate orchid]

is there anything funny?

[sly crescent]

Yeah

[delicate orchid]

please tell

[sly crescent]

You actually need (n²-n)/2 numbers to specify a Coxeter group of rank n

[delicate orchid]

explains why 3 numbers give tilings of a 2-dimensional surface

wiat does it

hold the phone 📱

[sly crescent]

These numbers correspond to the angles between facets of the fundamental domain

[delicate orchid]

I'm looking at the pictures on the wikipedia page, it seems like it determines how many triangles "touch" at a given corner

[sly crescent]

Yes

[delicate orchid]

right now it makes sense I think

there's 3 corners on a triangle, so you need 3 numbers

and it has that particular presentation because the product of two reflections looks like a rotation around a corner (I think?)

so if after n reflections you end up back at the start, (xy)^n = 1

[hidden wind]

aye

[delicate orchid]

woah you can REALLY generalise this.

I think I've just invented Tits buildings

[hidden wind]

i'm sure that exists already

[delicate orchid]

I did

[hidden wind]

a (l,m,n) triangle group has as a fundamental domain a Schwarz triangle with angles pi/l,pi/m/pi/n

[sly crescent]

We can represent Coxeter groups as labeled graphs by assigning each generating reflection a vertex. Then, the edge between them tells you the order of their product: if it’s 2, there is edge, if it’s 3, there is an unlabeled edge, and if it’s more than 3, there is an edge which is labeled with that value.

These are called Coxeter-Dynkin diagrams

[delicate orchid]

I know about them cause of rep theory

[hidden wind]

i see them all over the place but haven't bothered trying to make sense of them

[delicate orchid]

thinking about the generalisation here, if we just look at the simplicial complex consisting of an n-simplex but with the actual n-simplex removed, then you have (n+1) reflections as previously stated, and then the order of the product in the corrisponding Weyl group is the number of ``hollow" simplices around each 1-simplex I think? The trick is then trying to move this to general complexes

at least I have very good motivation to care about tits buildings now

[hidden wind]

are you familiar with the geometry side of things?

[sly crescent]

Here’s something cursed you can do: if you interpret the edge labels as just the fraction of pi that the angle forms, there’s nothing stopping you from using fractional labels.

[hidden wind]

actually those appeared when i was trying to make sense of the relation between (2,3,infty) and (infty,infty,infty) and i noped out

[sly crescent]

These don’t correspond nicely to presentations, though.

[delicate orchid]

irrational labels

or wait would they just be (\infty, \infty, \infty)

[hidden wind]

one thing i noticed is that a (infty,infty,infty) triangle can be formed by gluing together six (2,3,infty) triangles but i'm not sure how that correspond to the algebra side of things

[hidden wind]

there is a really nice thing here actually

[delicate orchid]

that's a good observation

algebraic "gluing" is a quotient (more generally, a colimit)

[sly crescent]

Here are the 5 Coxeter-Dynkin diagrams that generate tetrahedral symmetry

[hidden wind]

http://reu.dimacs.rutgers.edu/~ks1613/documents/series.pdf

[delicate orchid]

this is SO cursed

[sly crescent]

That means (∞,∞,∞) is an index-6 subgroup of (2,3 ∞)

[hidden wind]

that is what i was suspecting but i'm not sure how to show it formally 😭

[hidden wind]

^

teehee

[delicate orchid]

if ur a geometer u can just say "glue" and it'll be fine

cause that's all u mfs do anyway

[hidden wind]

ok i'll do just that, thanku for the pro tip

though i desperately need to actually learn elementary group theory

[sly crescent]

You can also label the vertices of Coxeter-Dynkin diagrams to represent uniform polytopes.

[hidden wind]

no time atm though, bachelor thesis due in 4 days and i've only written like 20% of what i intend (after cutting down on what i wanted to include drastically)

[delicate orchid]

since you're working with presentations I'll give you this tip: if you have a presentation <X|R> of a group G and <X|S> is a presentation of a normal subgroup in the same generator set, then G/N = <X|R union S>

[hidden wind]

uwu

[delicate orchid]

alternatively, if you want to add a set of relators S, you take the normal closure of S and then take the quotient

[hidden wind]

thanku thanku very much

[delicate orchid]

this is how everyone should think about quotients

[sly crescent]

I don’t entirely remember how this works, though

[hidden wind]

also this article is amazing (though i've only taken the time to understand the first half)

[delicate orchid]

Mecejide I've just noticed ur pfp is E_8

that's very fitting

[sly crescent]

No it isn’t

[delicate orchid]

it looks EXACLTY like E_8

[sly crescent]

It’s H_4

[delicate orchid]

three rings of thangs

oh nvm there's 4

[winged void]

I have question

How is orbit of a set is defined

[delicate bloom]

union of the orbits of the elements I'd guess

[sly crescent]

Wythoff Construction

[winged void]

I see

[winged void]

Can we use the orbit stabilizer theroem

Orem

If the stabiliser is a set

Not an element

[delicate orchid]

if your group is acting on X but you care about orbits on subsets of X, you can extend the action to a group action on the power set of X

then use orbit stabiliser there

[hidden wind]

group actions are very general, but while this indeed is possible, i suspect you're overcomplicating things here Mootje

though actually sometimes "overcomplicating" things yields lovely new insights mwahaha

[winged void]

Thanks

Idk if someone wants to check my prove

Cause I’m using matrices

@hidden wind

[hidden wind]

hi

[winged void]

Hi do you want maybe to check my prove

If possible

Almost finished

Please

[hidden wind]

just send it

me or someone else may check it

[winged void]

Sure give me 10 min because did not finish it

I have a long one

🥲

[delicate orchid]

it may even be a subgroup

[south patrol]

And elements are usually sets anyway

[winged void]

this is what

i did

\begin{equation*}
G := {g \in SO(4)|g(H) = H}
\end{equation*}
prove that $# G = 192$
\begin{proof}
First I'm going to find the number of elements of the stabilizer in $K_1$ since that's easier and then I'm going to prove that the action is transitive. To find the number of elements of stabilizer we have to follow the following steps. We know that $K_1$ has the following form:
\begin{equation*}
K_1 = {(x,y,z,w) \in H | x = 1}
\end{equation*}
This means that an element from $K_1$ must have the following form: $(1,y,z,w)$. However, we should note that the orthogonal group of $SO(4)$ is defined as follows:
\begin{equation*}
SO(4) = {g\in GL_4(\mathbb{R}) \mid g^{t}g = 1, det(g) = 1}
\end{equation*}
We further know that $SO(4)$ can be seen as the rotation around the origin of our four-dimensional cube. Since that element $x$ must be equal to one, the matrix rotation must look like this:
\begin{equation*}
\begin{pmatrix}
1 & W \
W^T & A
\end{pmatrix}
\end{equation*}
The above matrix is a $4 \times 4$ matrices. Where $W, W^t \in R^3$ and note that $A$ is a matrix of the form $A_{3\times 3}$. We further know that $g^tg$ must be equal to the identity and so this means that $WW^T = 0 $ This is of course only possible if $W = 0$ and therefore also $W = 0$. This means that there remains only matrix $A$ and this is a $3 \times 3$ matrix. so then for the rotation for the three dimensional cube we get the following matrix:
\begin{equation*}
\begin{pmatrix}
1 & 0 & 0 & 0 \
0 & \
0 & & SO(3)\
0 &
\end{pmatrix}
\end{equation*}
From here we just need to find the symmetry of $SO(3)$ and we know that they are equal to 24 so this means that the stabilizers of our $K_1$ is equal to 24.
\end{proof}

[cloud walrus]

Mootje

[winged void]

im not really sure if that is good way

and the reason that i did that is the following

give me asecond

[winged void]

i send it

Let $H \subset \mathbb{R}^4$ be the 4-dimensional cube given by
${(x, y, z, w) \in \mathbb{R}^4 \mid -1 \leq x, y, z, w \leq 1}$
(so the four coordinates lie between $-1$ and $1$).
We define
$\text{SO}(4) = {g \in \text{GL}_4(\mathbb{R}) \mid g^tg = 1, \det(g) = 1}$.
The group $\text{SO}(4)$ can be understood as rotations around the origin of 4-dimensional Euclidean space. Each such rotation is given by choosing two planes $V, W \subset \mathbb{R}^4$ and two angles $\theta, \psi$, where the planes are perpendicular to each other, and then rotating by an angle $\theta$ in $V$, and an angle $\psi$ in $W$. For example,
[
\begin{pmatrix}
\cos(\theta) & -\sin(\theta) & 0 & 0 \
\sin(\theta) & \cos(\theta) & 0 & 0 \
0 & 0 & \cos(\psi) & -\sin(\psi) \
0 & 0 & \sin(\psi) & \cos(\psi)
\end{pmatrix}
]
is such a rotation, where we rotate by an angle $\theta$ in the $x, y$-plane, and by an angle $\psi$ in the $z, w$-plane.
The rotation group of $H$ is given by
$G := {g \in \text{SO}(4) \mid g(H) = H}$
as a subgroup of $\text{SO}(4)$.
Problem: Prove that $#G = 192$.
Hint: The group $G$ acts on the "sides" of $H$. These sides are themselves 3-dimensional cubes, given by
\begin{align*}
K1 &= {(x, y, z, w) \in H \mid x = 1} \
K2 &= {(x, y, z, w) \in H \mid y = 1} \
K3 &= {(x, y, z, w) \in H \mid z = 1} \
K4 &= {(x, y, z, w) \in H \mid w = 1} \
K5 &= {(x, y, z, w) \in H \mid x = -1} \
K6 &= {(x, y, z, w) \in H \mid y = -1} \
K7 &= {(x, y, z, w) \in H \mid z = -1} \
K8 &= {(x, y, z, w) \in H \mid w = -1}
\end{align*}
Prove that this action is transitive, determine the stabilizer of $K1$, and conclude using a theorem we have seen in class.

[cloud walrus]

Mootje

[winged void]

this was the question

[dull ginkgo]

The only rotations that can map a plane to itself is quarter revolution rotations. Each plane has 3 faces / pair of orthogonal planes to it

[winged void]

what do you mean

i do not get it

[dull ginkgo]

I am sorta talking to myself, sorry

[winged void]

oh do not worry

do you have time to check my proof

of finding stabilisator

[dull ginkgo]

Let me check

[winged void]

sure

its the first image

the second image is the question

[dull ginkgo]

Idk seems fine. I don’t know the required level of rigor for your class

My psycho way of doing things is to let it act on {0,1}^4

[winged void]

the thing is im not really sure if W W^t

are in a good place

and the second thing that im not sure about if WW^t = 0 then W and W ^T must be 0

[dull ginkgo]

Sounds right

MM^T must be 0 as a whole matrix

Consider the image of (1,0,0,0) must be fixed

Or at least to one of the other verticies

And the transpose from there

I shouldn’t be using discord at a red light brb

[delicate orchid]

Oh wait. This is just the standard rep of S_4

[winged void]

so actually i should have written from her that in place of W^t mayb (1,0,0,0)

[dull ginkgo]

(1,0,0,0) must be mapped to one of the verticies of the cube where x = 1

[winged void]

sure but i do not really get what you want to say here

i mean sure the thing that we only know is that the following must be true

\begin(pmatrix)
1 & W \
0 & A \
0 & \
0 & \
\end{pmatrix}

this is what you mean

maybe

and what do you thing @delicate orchid

[delicate orchid]

honestly, mootje? I like your proof. It's a good approach

it's just very explicit

[winged void]

thanks but is it good

is it a good approach

and now im struggling with now proving transitivity

or do you have some changes

[delicate orchid]

I just... I just don't believe it's 192 though? I just don't

The hyperoctahedral group for $n=4$ is $S_4 \wr C_2$ which has order $24^2 \times 2$ which is like, 1152

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

sure, the fact that they're special orthongal gets rid of that factor of 2, but it's still 24^2

[winged void]

hahaahah

[delicate orchid]

I just don't even know what action is on the hyper cube here?

[winged void]

well i mena i just proved that the element in stabilisator are 24

[delicate orchid]

are we allowed to pick up and rotate individual cubes?

[winged void]

and you have agreed with my prove

[delicate orchid]

I agree with the approach

[winged void]

but well i still did not find the orbit to be equal to 8

thus

so we do not know if its correct

i just found that the stab equal to 24

[delicate orchid]

because the size of this group just for the cube is absolutely massive

well no we're not allowed to do that because we're in SO(4) so we can only rotate

[winged void]

true

[dim widget]

Isn’t it though?

[delicate orchid]

how?

[dim widget]

I think this proof works

[delicate orchid]

every possible thing I've come up with is massive

[delicate orchid]

I never doubted that

I just don't believe it

[dim widget]

Probably you have written down things that are not rigid symmetries of the hypercube which preserve orientation

[delicate orchid]

I haven't written anything down this is pure visualisation

[dim widget]

Hmm well I think for visual proofs this approach is pretty good

Clearly the symmetries act transitively on the faces

[delicate orchid]

yeah which is why I like it. Work with the rotations directly

[winged void]

well is my proof correct btw

[dim widget]

So then we just have to compute the stabilizer of one face which is a vanilla flavored cube

[winged void]

or do you find it rigorous

[dim widget]

and then that can be calculated to be S_4

[dim widget]

can you post it? I have not read it

[winged void]

sure give me a second

[dull ginkgo]

SO(4) acts on verticies of the hypercube transitively, which corresponds to {0,1}^4. The orbit is therefore 2^4 = 16. Now it’s asking how many rotation stabilize a vertex, so for example consider the vertex (1,0,0,0). The vertex must be fixed here, so the matrix in SO(4) must have its first row and column be (1,0,0,0). Thus the remaining bit is the lower 3x3 matrix?

[winged void]

i will post new version

[dull ginkgo]

Of which must permute the remaining verticies?

[winged void]

\begin{equation*}
G := {g \in SO(4)|g(H) = H}
\end{equation*}
prove that $# G = 192$
\begin{proof}
First I'm going to find the number of elements of the stabilizer in $K_1$ since that's easier and then I'm going to prove that the action is transitive. To find the number of elements of stabilizer we have to follow the following steps. We know that $K_1$ has the following form:
\begin{equation*}
K_1 = {(x,y,z,w) \in H | x = 1}
\end{equation*}
This means that an element from $K_1$ must have the following form: $(1,y,z,w)$. However, we should note that the orthogonal group of $SO(4)$ is defined as follows:
\begin{equation*}
SO(4) = {g\in GL_4(\mathbb{R}) \mid g^{t}g = 1, det(g) = 1}
\end{equation*}
We further know that $SO(4)$ can be seen as the rotation around the origin of our four-dimensional cube. Since that element $x$ must be equal to one, the matrix rotation must look like this:
\begin{equation*}
\begin{pmatrix}
1 & W \
W^T & A
\end{pmatrix}
\end{equation*}
The above matrix is a $4 \times 4$ matrices. Where $W, W^t \in R^3$ and note that $A$ is a matrix of the form $A_{3\times 3}$. We further know that $g^tg$ must be equal to the identity and so this means that $WW^T = 0 $ This is of course only possible if $W = 0$ and therefore also $W = 0$. This means that there remains only matrix $A$ and this is a $3 \times 3$ matrix. so then for the rotation for the three dimensional cube we get the following matrix:
\begin{equation*}
\begin{pmatrix}
1 & 0 & 0 & 0 \
0 & \
0 & & SO(3)\
0 &
\end{pmatrix}
\end{equation*}
The reason that it must be a $SO(3)$ is very simple because we know one more thing that the determinant of the matrix must be equal to $1$ so this can either be the entire diagonal equal to $1$ or with the development of the minors we need to get determinant $1$ \end{proof}

[delicate orchid]

oh lord please don't post all that again

tteg can u just scroll up?

[cloud walrus]

Mootje

[winged void]

oh sorry

[dim widget]

no

[winged void]

well im going to delete it

[delicate orchid]

it's ok it's just very long

[winged void]

oh sure

[winged void]

yes but this my proof

i did not yet prove teransitivity

and idk how

but i found stabilsator is easier

[delicate orchid]

transitivity is pretty clear

[winged void]

well i mean if its clear

than we get the orbit

but idk how the orbit must be 8 thought

[delicate orchid]

to move one face to an opposite face, pick any other face's midpoint as your axis and rotate by 180 through it, for adjacent faces do the same thing but at a corner by uhh probably 90?

[tribal moss]

In a hypercube, shouldn't the vertices be (±1,¯1,±1,±1)? Things like (1,0,0,0) would be the vertices of, um, a hyper-ocathedron instead.

[delicate orchid]

wait is the cube centred at the origin?

[dim widget]

I did not understand this proof

[delicate orchid]

u know what. It better be

[dim widget]

yes

[delicate orchid]

good

[winged void]

well you need to read the question

[tribal moss]

I assumed so, since the natural action of SO(4) fixes the origin.

[winged void]

i will tag the question

[dim widget]

I know the question

[delicate orchid]

true!

[winged void]

this

[dim widget]

I don’t understand how you show that the upper left hand entry of the matrix is 1

[winged void]

read this

[rocky cloak]

Should be 2^4 * 4! I think. So if you only consider rotations not reflections that's 2^3 * 4! = 192

https://en.m.wikipedia.org/wiki/Hyperoctahedral_group

[delicate orchid]

that is far too much text for a problem I completely know about

[delicate orchid]

am I just stupid. Oml did I get the wreath product backwards

[winged void]

hahahah i know

[delicate orchid]

I got the wreath product backwards chat

[winged void]

but @delicate orchid how would you prove that its transitive

or the orb is equal to 8

[delicate orchid]

I just said

oh wait what about the secret third kind of face

[winged void]

yes but this more of an intuition

[delicate orchid]

it really isn't

those are explicit elements of SO(4)

[rocky cloak]

Guess you can just explicitly argue that things like
(x, y, z, w) -> (z, x, y, w) and so on are symmetries

[delicate orchid]

you have to be careful with orientation preserving

[winged void]

that is true

[coral spindle]

Ought just to be the even permutations then, right?

[tribal moss]

Then you have A4.

[delicate orchid]

but A_4 acting transitively is... obvious?

yeah

we're fine

still cannot believe I remember the wreath product wrong

[winged void]

well the element of A4 are 12

[delicate orchid]

one of my worst blunders

[winged void]

not 8

so then we have that its not 12 * 24 does not equal to 192

[delicate orchid]

yeah, and we're acting on the basis vectors

[dull ginkgo]

So brain missed the centered at the origin part

[winged void]

except if the 24 that i found is wrong

[rocky cloak]

A face is just what you get if you fix one cordinate, so this together with flipping two signs shows you you can move from any face to any other

[delicate orchid]

no it isn't, we're saying that even permutations of the coordinate axes are in SO(4), and fix the cube

[tribal moss]

12×16 is 192, though, and the hypercube has 16 pairs of opposite vertices to preserve.

[delicate orchid]

and this action is transitive

[tribal moss]

Wait, no.

I off-by-one counted a dimension somewhere.

[winged void]

however its suppose to be 24

im sure about that

since you also agreed to my prove

and sure that the order must be 8\

[dim widget]

Why not just look at the action on vectors of length one with only one nonzero entry

[delicate orchid]

what? I never said it wasn't?!?!

[winged void]

well you said it is not

[tribal moss]

(can someone state in a single line what it is we're counting?)

[dim widget]

Then you have the elements of SO(4) of the form diag(-1,-1, 1, 1) and all of the even permutations, and the group they generate

[delicate orchid]

we're showing that the action is transitive I couldn't care less about the size of the stabiliser

and where?

[dim widget]

This acts transitively

[rocky cloak]

Rotational symmetries of tesseract

[winged void]

do not worry i read it wrong

[tribal moss]

You meant diag(-1,-1,1,1), I hope?

[dim widget]

Yes

[winged void]

sure

[dim widget]

Okay so that shows the action is transitive

Now the stabilizers of a face leave the midpoint of the face fixed

So wolog (1, 0, 0, 0) is fixed

[tribal moss]

Do we already have a known solid argument for 192, and we're looking for alternative ways to get the same?

[dim widget]

so the matrix is block upper triangular and orthogonal, so it is a block matrix of the form you say

[winged void]

this is what we are suppose to prove

[dim widget]

Yes

[winged void]

its true

[dim widget]

Or rather

[tribal moss]

That was ... somewhat conflicting replies.

[winged void]

its given

[delicate orchid]

wait really?

[dim widget]

We have a sketch and are helping mootje fill in the details

[delicate orchid]

oh ok!

[dim widget]

The sketch is to look at the stabilizer of a face and use orbit stabilizer

[winged void]

but i assume that the amount of elements here must equal to 8

[tribal moss]

(My first reaction would be to say, we can permute the axes in 4! ways and for each of them choose signs in the permutation matrix in 2^4 ways, but half of those have the wrong determinant, so 4!·2^4/2 = 192).

[winged void]

i might be wrong though

[dim widget]

No

[delicate orchid]

I'm writing up a full proof of this as we speak

[dim widget]

There are big stabilizers

[delicate orchid]

and is what I had in my mind and would've said if I had not gotten the wreath product backwards!

[dim widget]

But it doesn’t matter because we can compute the stabilizers later

[winged void]

sure

[delicate orchid]

actually it doesn't matter tropo said it better than me lol

[winged void]

the orbit should be 8

[delicate orchid]

nobody is saying it isn't 8

please stop restating this

[winged void]

well sorry

now im just sharing information

well the thing i do not get how to prove that

i mean sure we prove transitivity

and we know that transititvity implies

that there exist one orbit

but the thing is we have also find a stabilsator of a subset

not the whole set

[dim widget]

Oh my

You asked for help proving that the group acts transitively on the faces

[delicate orchid]

here's what I had anyway:
Consider the set of reflections r_i sending e_i -> -e_i in R^4 and the group they generate. This group acts transitively on the faces of the hypercube. These have determinant -1 as linear transformations and are in O(4). The product of two reflections r_ir_j has determinant (-1)^2 = 1, and is thus in SO(4). Furthermore this product is a rotation, shown by an elementary examination of whatever dihedral group r_i, r_j generate in the plane <e_i, e_j>. These account for all rotations of the hypercube because \todo{explain why}

[dim widget]

That’s why we’re helping

[winged void]

sure i appreciate as well

[delicate orchid]

I don't mean to be rude but are you serious?

[winged void]

but i mean like if we prove that transitive then we must at the same time know the orbit

[delicate orchid]

are you actually serious? Maybe mention this 20 fuckin minutes ago?

I'm out

[winged void]

hahahah

well just forget about it

i mean i got confused

to be fair

[tribal moss]

To say "the action is transitive" nothing more or less than a fancy word for "there is a single orbit containing all the things".

[winged void]

i agree with that

and i was mentioning that like 10000 times

the thing that went wrong a lot is speaking that i missed the part

that we are suppose to prove

cuz im writing in latex in overleaf and coming back

so i got confused

and then Wew lads got angry

on me

[delicate orchid]

I'm not really angry

[winged void]

sorry

[delicate orchid]

it's alright but in future just make it clear when a problem has actually been solved so someone else doesn't get trolled online

[winged void]

sure

[delicate orchid]

ok but now I am actually curious what this other problem is

[whole basalt]

So I’ve been thinking about elliptic curve cryptography recently and the possibility of incorporating it into my abstract algebra class. I’m wondering — do people use them much over finite fields of prime-power order rather than just prime order?

I’m looking to use applications to anchor why we study some of the things we do, so for example I’m going to be using the Rubik’s cube as an anchor for group theory.. I figured elliptic curve cryptography could be good for finite fields and other topics, especially if they can be analyzed over extension fields.

[prisma ibex]

yeah there are some examples like this. For example people will often work in characteristic 2 in order to exploit binary representations of elements, and then the relevant curves are defined over finite fields F_{2^n}.

There are other examples like the FourQ curve which is defined over F_{p^2} for the Mersenne prime p=2^{127}-1

[south patrol]

what about the fargues fontaine curve

[whole basalt]

Doesn’t some weird stuff happen when you’re in characteristic 2 though?

[delicate bloom]

the weierstrass normal form won't work, you'll have to use a slightly different form of that. Probably other problems too lol

appears discord is using an elliptic curve digital signature algorithm currently, could look into this one specifically if you want feet-on-the-floor motivation:

[primal beacon]

the specific finite field features may be a small bit complicated if you’re against people just starting algebra and wont be specifically focused on the finite field stuff, still it is doable. I suggest you to check about error correcting codes, its a good bunch of linear algebra over finite fields and it uses in an essential way the basic knowledge of finite fields. Another suggestion is LFSR’s which are pseudo random generators which again to prove the main prosperties use the basic finite field properties in an essential way

both these thing are much more elementary than elliptic curves (still very very interesting and used in practice)

[hidden wind]

my supervisor telling me to keep on neglecting abstract algebra like a chad

[ashen heron]

can someone verify this for me

[cloud walrus]

pika

[whole basalt]

I'll look into those — could be very useful, thank you!

[void cosmos]

Galois Field computations used in AES to compute inverses

specifically GF(2^8) iirc

[whole basalt]

AES?

[void cosmos]

yeah

[whole basalt]

What's AES :V

[void cosmos]

advanced encryption standard, an encryption algorithm

[whole basalt]

Oh interesting

[void cosmos]

yeah it's pretty much used in most wireless security things

like WPA2 ( the thing ur using for wifi security )

[whole basalt]

To be honest I'm not sure if I'm gonna go this far in my abstract algebra class next time I teach it, but I'm considering my options

[void cosmos]

and so on

I think elliptic curve cryptography is a very good suggestion imo

[whole basalt]

I was trying to get to the quintic and make the topics as accessible as possible, but I think it went over some of their heads :/

Because I thought the unsolvability of the quintic was a compelling enough focal point

[void cosmos]

you mean prove the unsolvability of the quintic?

[whole basalt]

Yeah

[void cosmos]

yeah especially cuz it's pretty easy to motivate and has a cool story

i agree

[whole basalt]

So much of the field theory links right back to stuff they've learned in high school

[void cosmos]

what about maybe t he trisecting problem

trisecting an angle iirc

or would that be boring

[whole basalt]

And answers fundamental questions like "how do we invent new numbers like i or √2"

Hmm, I didn't go into geometric constructibility

[void cosmos]

yeah

Ig the unsolvability of the quintic is the coolest motivator

in my "student" opinion haha

[whole basalt]

Yeah, but my students had a lot of trouble keeping definitions etc straight throughout the semester

Like a student told me they needed to look up what a field was every time they were asked about it

[void cosmos]

yeah i see

yeah ig it would be tough to like

introduce galois theory

more definitions regarding types of field extensions and so on

imo

[mighty kiln]

Why not introduce covering theory

[void cosmos]

!!

you mean

the covering space galois correspdence ?

[whole basalt]

I think they were having trouble seeing the big picture

[void cosmos]

then ig ur thinking with like

[void cosmos]

this

I think this is the best way to go imo

[mighty kiln]

Group applications

[void cosmos]

u were talking about that right?

I just learnt that like some months ago

when learning aglebraic topology

and it's crazy

you can literally solve problems regarding free groups with just drawing some covering space of S^1 wedge S^1

it's so beautiful

[whole basalt]

For group applications, I was thinking of using the Rubik's cube as the grand motivator

[void cosmos]

yeah that's pretty cool too

or maybe chemistry too

spectroscopy ( things i dont know about :V )

[whole basalt]

I don't know much about the chemistry side of things other than like ... molecules have symmetries 🤷‍♂️

[mighty kiln]

Is there quantum involved

[void cosmos]

yeah definitely

[void cosmos]

omg what if

you

motivate symmetries a ton

then like

do the boring stuff ( the formal definitions, weird examples etc,..)

and then do cayley's theorem

boom

lmfao

as in like

every group is basically some subgroup of symmetry group

isn't that crazy

[dull ginkgo]

no, let the group act on the set of itself by multiplication

Now the Cayley theorems on the other hand

[void cosmos]

yeahh

[dull ginkgo]

,rotate

[void cosmos]

ew

[dull ginkgo]

I am going to try to do all three of those

The first one I assume is like multiplying p_3 and p_2 and sieving out the extra repeating factors through power sums

[void cosmos]

im not thinking into that

hahahaha

i dont wanna know

i hate math

[hidden wind]

symmetric polynomials are great

[dull ginkgo]

I assume I have to multiply p_2 * p_3 and sieve out terms I don’t want

Anything with a x_k^2 in it

[tardy hedge]

How do i find a cyclic subgroup of order 8 in the group of units of Z32 without just guess and check

Im just calculating different things but seems inefficient

[mighty kiln]

Do you what the group of units of Z/n looks like

[tardy hedge]

Integers less than n relatively prime to n

[mighty kiln]

What about the group structure?

[tardy hedge]

I do not know

Sometimes cyclic sometimes not

I dont think ive learned theorems about group structure of that. Maybe i learn it later?

I have not learned the structure thm of finite abelin groups if thats what ur referring to

That comes later in my book

[loud merlin]

I think Arki just ask what is the composition law on (Z/n)^x

[mighty kiln]

No

[tardy hedge]

Composition law?

[mighty kiln]

Then maybe they just expected you to trial and error

[tardy hedge]

Probably, it is only the first question so its probably meant to be eaay

Easy

[loud merlin]

wdym by group structure then

[tardy hedge]

It was just an application of some theorem about if G = HK and some other hypothesis, then G is direct product of H and K

[dull ginkgo]

Chinese remainder theorem to split it into a direct product

Then observing those rings, seeing if you can find a prime (which is cyclic of order one less than the prime)

Oh wait it’s 32

Z/2^5Z

[tardy hedge]

@dull ginkgo that is ahead of where im at rn

I believe

[dull ginkgo]

It doesn’t even apply and I was about to assert that there’s a primitive root generator but it’s a power of 2 so that doesn’t even happen lmao

[tardy hedge]

Oh ok

[dull ginkgo]

It’s the multiplicative group of elements coprime to 2 less than 32

I.e odd numbers between 0 and 32 (both exclusive)

Actually while I’m at it I might aswell try to find the structure of Z_2^k

(2n + 1)(2m + 1) = 4nm + 2(n + m) + 1

Let’s say we are in (Z_2^k)^x
(2^a + 1)(2^b + 1) = 2^(a + b) + (2^a + 2^b) + 1

[dull ginkgo]

Just immediately try finding the order of 3

[tardy hedge]

@dull ginkgo you’re cracked

[dull ginkgo]

Thanks to this problem I realized that multiplicative groups mod n are fucked

You know, prime powers could be nice and all be cyclic y’know

But nah, powers of two have to be fucking “special”

fuck.

[dim widget]

For the record $(Z/N)^* \cong \prod_{p|n} (Z/p^{n_p})^$ where $n_p$ is the largest power of p which divides N. Then $(Z/p^n)^ \cong (Z/p)^* \times Z/p^{n-1}$ when p is odd, and $Z/2 \times Z/2^{n-2}$ otherwise

[cloud walrus]

Math_Discord_Final_Girl

[dull ginkgo]

I thought it’s cyclic

Wait no I am an idiot

[dim widget]

except for 2 they are cyclic

but the powers of 2 ones are not

Unless n is small

[dull ginkgo]

Yeah the higher powers of 2 past 4 not being cyclic is stupid

I think it’s a product of cyclic of order 2 corresponding to <-1>

and <3> which has order 2^(n-1)

[dim widget]

you think correctly

[dim widget]

yep

[dull ginkgo]

Thanks to Kiand’s exercise ye

[dim widget]

In general the additive part of Z/p^n * is generated by 1 + p

the rest is the elements from 1 to p-1, or their teichmuller lifts if you want to be fancy

[dull ginkgo]

Add it to my very few uses for CRT

• fully describing multiplicative groups mod b
• Artinian commutative rings are product of finitely many local rings assuming choice so the maximal exist lol
• Independence of characters over Fields (thus integral domains)

Idk what else it’s used for that don’t go down to those 3 cases though I did use it to make the world’s double indexing imaginable to torture my friend

[glad osprey]

They're not cyclic because the Carmichael function of 2^n is less than phi(2^n) for n>2, right? ie. no elements have order 2^n

[dull ginkgo]

[dull ginkgo]

I mean that’s kind of the byproduct I guess lmao

[glad osprey]

lambda(2^n) < phi(2^n) because Z/2^n is not cyclic?

[dim widget]

that’s the definition yes

[glad osprey]

I guess you could argue the implication goes both ways. But proving lambda(2^n) < phi(2^n) is relatively straightforward number theory, so that's the direction that makes most sense to me

https://en.wikipedia.org/wiki/Carmichael_function#Sharpening_the_result_for_higher_powers_of_two

[dull ginkgo]

So this is how Jacobson tells the reader how to do this problem

[dim wagon]

was looking through my course outline, am i correct that the content coverage is on the low-end? It's a semester long course, 3 hour lectures/week for 13 weeks

[tardy hedge]

Yeah thats like one chapter

1-2 chapters of a textbook

[dull ginkgo]

This feels extremely like overcomplicated?

Let $s_k = \sum_{n = 1}^{N}{x_n^k}$. Now $f(x) = \prod_{n = 1}^{N}{(x - x_n)} = \sum_{n = 1}^{N}{(-1)^{N - n}p_{N - n}x^n}}$ \
Thus $f(x_n)x_n^m= 0$ for $1 \leq n \leq N$. So $0 = \sum_{n = 1}^{N}{f(x_n)} = \sum_{n = 1}^{N}{(-1)^{N - n}p_{N - n}s_{n+m}}$

[cloud walrus]

THE TUBE
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dull ginkgo]

Let $s_k = \sum_{n = 1}^{N}{x_n^k}$. Now $f(x) = \prod_{n = 1}^{N}{(x - x_n)} = \sum_{n = 1}^{N}{(-1)^{N - n}p_{N - n}x^n}}$ Thus $f(x_n)x_n^m= 0$ for $1 \leq n \leq N$. So $0 = \sum_{n = 1}^{N}{f(x_n)x_n^m} = \sum_{n = 1}^{N}{(-1)^{N - n}p_{N - n}s_{n+m}}$

slightly modified

[cloud walrus]

THE TUBE
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[next obsidian]

lol bro has compile error

chmowned

[dull ginkgo]

either way does this like 10 second solution work lmao

[next obsidian]

I didn't read that shit

I just made fun of u for it

[dull ginkgo]

I came up with that in like a fucking TENTH of the time it took to read the goddamn hint Jacobson gave which made 0 sense

[next obsidian]

then it's wrong

cuz Jacobson is 300x the man you'll ever be

[dull ginkgo]

I don't see a single goddamn issue with it

[next obsidian]

latex it properly

and then I"ll read it

[dull ginkgo]

Let $s_k = \sum_{n = 1}^{N}{x_n^k}$. Now $f(x) = \prod_{n = 1}^{N}{(x - x_n)} = \sum_{n = 1}^{N}{(-1)^{N - n}p_{N - n}x^n}$ Thus $f(x_n)x_n^m= 0$ for $1 \leq n \leq N$. So $0 = \sum_{n = 1}^{N}{f(x_n)x_n^m} = \sum_{n = 1}^{N}{(-1)^{N - n}p_{N - n}s_{n+m}}$

[next obsidian]

chmowned

[cloud walrus]

THE TUBE

[next obsidian]

How tf you know the third =

wait I think I get it

that's kinda obvious

[dull ginkgo]

YEAH

[next obsidian]

look at this guy tryna argue it rather than say it's obvious

waht a dummt

I'm so smart

[dull ginkgo]

tldr exchanging the summation signs

[next obsidian]

I think that makes sense ngl

You might be the goat

Hi walter and bubu

[summer path]

hi chuchu

[next obsidian]

I love u bubu

I love u walter

[agile burrow]

Love u too chmonkey

Sorry, I was putting my toaster away

[crystal vale]

What is meant by HOM(A,B) = B^A I think they mean cardinality of A and B

[lusty marlin]

B^A is commonly used in set theory to denote the set of functions from A to B

[crystal vale]

Okay, thank you

[manic brook]

i might just be blanking in this question but how do we extend the automorphism from F to F(u)
i think i'm not seeing the powers of u getting fixed when they are adjoined, because the rest of it falls into place just from the problem statement

[mighty kiln]

A field automorphism of K is given

It is known that F and u are fixed

Show that this automorphism fixes F(u)

[manic brook]

any element in $F(u)$ is a linear combination of powers of u with coefficients in $F$, since automorphisms are isomorphisms then we can break the polynomial into additions and multiplications that we know are fixed since coefficients are in $F$ which are already fixed, and powers of $u$ can be factored as $\sigma(u^n) = (\sigma(u))^n$?

[cloud walrus]

HyacinthDroplet

[crude flame]

Isn't the answer 6?

[delicate bloom]

should be the lcm of the orders of each of the elements individually, can you show your work

[crude flame]

$lcm(o(7),o(2),o(-k),o(-1))=lcm (6,1,2,2)=6$

[cloud walrus]

Bilal ns

[chilly ocean]

I would double check all those orders you got

6*7 isn’t 0 in Z10

[kind temple]

the order of 2 in Z_6 is 3 wrong as well

[rocky cloak]

and -k in Q8

[crude flame]

Sorry ,my mistake . it's number of elements of order 7 and 2 in $Z_{10}$ and $Z_6$

[cloud walrus]

Bilal ns

[rocky cloak]

? What was meant to be that number?

[crude flame]

$lcm(o(7),o(2),o(-k),o(-1))=lcm (10,3,4,2)=60$

[cloud walrus]

Bilal ns

[rocky cloak]

Looks good 👍

[warm ember]

is every separable extension galois

then why does galois closure exist

Given a separable extension K′ of K, a Galois closure L of K′ is a type of splitting field, and also a Galois extension of K containing K′ that is minimal, in an obvious sense.

[dim widget]

no

[dim widget]

to study general fields using Galois theory

[warm ember]

huh

[delicate orchid]

No, normal separable fields are Galois

[warm ember]

normal?

[dim widget]

what about abnormal ones?

[delicate orchid]

Bizarre extensions…

[delicate orchid]

A normal extension is an extension such that if one of the roots of a polynomial is in that extension, all of them are

I think. Correct me if I’m wrong

[warm ember]

oh

ok thanks

also the minimal polynomial of any algebraic number is separable right

how would you prove this

or is this not right

[sonic coral]

isn’t this just the same definition as a splitting field

[warm ember]

at least this should be true in Q right

[rotund aurora]

all irreducible polynomials over a field of char 0 are separable

[warm ember]

oh ok ty

[chilly radish]

No. A splitting field is for a specific polynomial. Here any polynomial with a root splits.

[warm ember]

o ig smth like (x-a)^p=x^p-a^p isnt separable in char p field

[chilly radish]

A splitting field is a normal extension though

Generally a normal extension is a splitting field for a family of polynomials

[sonic coral]

ah i see okay

[south patrol]

Though you can just multiply together all your polynomials right

In the finite case

[rocky cloak]

The condition that any irreducible polynomial is seperable is called being perfect.

Any characteristic 0 field is perfect, like croq said. And also any finite field, or algebraic extension of Fp.

So for a field to be not perfect it would have to at least contain Fp(t) for some p.

Simplest example being
x^p - t
over Fp(t), which is irreducible but not seperable.

[next obsidian]

Poly is not separable iff it and its minimal poly share a root.

This proves any irreducible poly over char 0 field is separable.

Also makes clear that irreducible poly is not separable iff derivative is 0

[south patrol]

this is a nice summary of it actually lol

like F_p(t) in some sense being universal for this

now is F_1 perfect

[rocky cloak]

According to Wikipedia, field extensions of F1 are always given by 'adjoining roots of unity' and corresponds to cyclic groups. So I guess that should mean they're seperable...

So I guess that would make F1 perfect

I guess one should take some alggeo stuff that only works for perfect fields and check what happens

[delicate orchid]

It’s perfect to me

[languid trellis]

It's perfect vacuously, as F_1 doesn't actually exist

[delicate orchid]

Wait you mean you’re extending by cyclic groups or the extenstion IS a cyclic group?

The latter feels wrong to me

[sonic coral]

aren’t fields defined to have different additive and multiplicative identities

so how would F_1 be a thing

[delicate orchid]

You don’t want to know

[dim widget]

it’s not a literal field with one element

it’s a metaphor which is quite complicated and the correct object probably hasn’t been defined yet

[delicate orchid]

It’s the thing you plug into, what I presume in generality are group schemes, to model what happens if you set q = 1 for formulas for things over F_q

[dim widget]

there’s a sense in which number theory looks like algebraic geometry of “curves” over some “field” and the field with one element is supposed to be that “field”

[delicate orchid]

This for instance turns GL(n, q) into S_n both globally and locally

Which is why I care

[languid trellis]

I feel like working with S_n is worse

[sonic coral]

what’s the typical progression after a galois theory course

[delicate orchid]

That’s because you’ve never had to compute the reps of either of them.

[agile burrow]

Vector spaces over F_1 are supposed to be finite sets?

[dim widget]

for many people it’s never using Galois theory again

[dim widget]

yes

[delicate orchid]

Think about it like this. You’ve done group actions, which are F_1-reps. But you have not done vector space reps

[languid trellis]

I know what reps are. 8 reps is a set

[sonic coral]

i really liked the class i just had on it though

[delicate orchid]

8? That’s cardio

[languid trellis]

Maybe for you

[rotund aurora]

what would even be a field extension of F1

oh wait jagr mentioned it above, didn't read all

[dim widget]

It should somehow be that there is a unique extension of size n for any n and all of them are the finite sets \mu_n

that’s my understanding

so that the maximal abelian extension of Q plays the role of \overline{F_p}(T)

[teal wind]

Thread

Does anyone understand what the klein group v4 is?

[rocky cloak]

Like Hom(F_(1^n), -) is the group scheme of nth roots of unity.

[rocky cloak]

From what I've run into they're pointed sets. And F1-linear maps are maps of pointed sets that satisfy the first isomorphism theorem.

[sonic coral]

the only group of order 4 which is not cyclic

[teal wind]

Does this definition make sense?

[coral spindle]

Yes

[warm ember]

o thanks

[teal wind]

Do you understand how I'd do part c here?

Do you understand this solution?

and it follows that because that, U_12 cannot be isomorphic to R_4.

[coral spindle]

Yes

[teal wind]

But then how did they figure out U_12 is not isomorphic to V_4?

They perform the mod based calculations

and they similarly found out that there is no element of U_12 whose square is not the identity...

But why? Why is it isomorphic to V if there is no element of U_12 whose square is not the identity

wait

[chilly radish]

Good morning algebra

[sonic coral]

all elements can’t have an order of 2

[delicate orchid]

just change it to all non-trivial

[chilly radish]

Sanity check time: The proof that a basis of a VS remains a basis under extension of scalars goes through for free modules right?

[rocky cloak]

Yes, tensor products commute with direct sums (as well as all colimits)

[rocky cloak]

Are you from Hawaii or something?

[chilly radish]

No just like to say good morning at odd times

[rocky cloak]

Always morning somewhere I guess

[celest furnace]

What if he just goes to bed late?

[rocky cloak]

Is morning relative to when you wake up, or relative to the day?

Is it always morning when one wakes up?

[south patrol]

Good morning

[south patrol]

The day I guess

[summer path]

Good meowning

[south patrol]

Though I feel like people usually use it to mean strictly before 12pm

though actually like it seems more controversial as to when it begins

When it begins seems more relative to the person ig and I think that is just a social thing

unless you want to say it is after dawn or smth lol

[summer path]

Morning to me is the time between waking up and eating "lunch" whatever that means

[south patrol]

So for me morning is about 20 minutes long

[summer path]

For me it's 5am-4pm

[rotund aurora]

How is 15:59 pm "morning"

[summer path]

Haven't eaten lunch yet, must be still morning

[next obsidian]

Hallo every-nyan

[dull ginkgo]

Do all adjunctions commute with colonists

[mighty kiln]

Left adjoint commute with colomits

[south patrol]

doesn't really make sense to say "adjunction" here, but rather "(in this case, left) adjoints"

[rocky cloak]

And right adjoint commute with limits

[south patrol]

this is formally dual (see there for more examples)

[summer path]

i can't tell if the misspellings here were intentional or not

(why are colonists commuting?)

[rocky cloak]

Presumably the colony is quite far away, so they have to commute to their job. And I guess they all work as adjuncts at the university

Those left anyway.

The right adjuncts are those that limit commuting by moving to the city

You sometimes see them pair up in formal duels, fighting each other

[dull ginkgo]

I need to prove that Z[X] satisfies the divisor chain condition

Here’s my thoughts chat

Assume we have a strict chain of divisors where a_(n+1)(X) | a_n(X)

Obviously deg_(a_n+1) <= deg(a_n)

The degree of a nonzero polynomial has to always be positive (or zero) so by pigeonhole the degrees stabilize to some n

[celest furnace]

I am a stern believer of it is when you wake up.

[dull ginkgo]

And if a(X) | b(X) and deg(a) = deg(b), then a(X) * c = b(X) for some constant c in Z

I am trying to invoke contradiction here

[rocky cloak]

Maybe think about a specific coefficients of a

[tardy hedge]

If G is not abelian, does there have to exist a subgroup of G that is not normal?
If all subgroups of G are normal that does not imply G is abelian does it?

[dull ginkgo]

When the chain’s degrees stabilize, each successive term differs by an integer factor. The constant coefficient epimorphism fixes Z and thus the image of the chain after it stabilizes, which would give a strict chain of divisors in Z, a contradiction

[dull ginkgo]

Nontrivial subgroup I take you mean

[tardy hedge]

yeah

[chilly ocean]

Why is k being coprime with |H| needed? I thought I proved it but I didn't need to use that hypothesis

If G/H = <gH> then take K = <g> and it's easy to see G = KH and H intersection K is trivial argue by Lagrange's theorem the order of H ∩ K divides both k and |H| so it must be 1. Then take the internal semidirect product

[rocky cloak]

Consider Q8 for example

[tardy hedge]

what is Q8?

[rocky cloak]

Quaternion group

[tardy hedge]

Oh I have not looked at that example before

[summer path]

quaquaquaquaternions

[rocky cloak]

It has elements
{±1, ±i, ±j, ±k}
with i^2 = j^2 = k^2 = ijk = -1

[rocky cloak]

You may want to double check why it's "easy to see"

Consider for example G = Z/4 and H = 2Z/4

[coral spindle]

It’s easy to see that every extension is split

[sonic coral]

do you know the second smallest group with that property

[rocky cloak]

https://planetmath.org/hamiltoniangroup

A group is Hamiltonian if and only if it is isomorphic to Q8×P for some torsion abelian group P that has no element of order 4.

So Q8 x C2 should be the next one

But in essence Q8 is kinda the only one

[sonic coral]

that’s interesting

thanks for that

[coral spindle]

Q8 is also the smallest group with a representation with a nontrivial real (or for that matter, rational) Schur index!

[delicate orchid]

that 2 dim sure is a doozy!

I presume it's that one anyway

[coral spindle]

Yeah that's right

The Schur index divides the degree, so in the first place it can only be the one of degree 2

[delicate orchid]

you have to be 4 dim over the reals

[coral spindle]

Precisely!

[delicate orchid]

because uhh something something Brauer group is C_2

[coral spindle]

Yes!

[delicate orchid]

I win?

I won

[coral spindle]

And guess what the endomorphism algebra of that real representation is?

[delicate orchid]

hmm...

K[x,y]/(x^2+y)

final offer

[coral spindle]

:(

Central simple real algebra, wew

[delicate orchid]

is it H perchance

[coral spindle]

It may be H perchance

[delicate orchid]

wow! almost like I didn't have too many choices!!

[coral spindle]

Ain't that cool?

[delicate orchid]

I mean this is the simplest possible case, what else is it going to be

what is cool is that this lets you differentiate between D_8 and Q_8 character theoretically

with some information on centraliser sizes

[coral spindle]

Yeah!!!

Because otherwise Thifford Cleory makes their character tables the same aaaa

I love character theory

[delicate orchid]

why are you using Thifford Cleory for such an assertion

bro forgot row orthogonality 😹

[tardy hedge]

what is character theory about?

[delicate orchid]

one ring isomorphism followed by combinatorics

[tardy hedge]

i see

what is higher level combinatorics like?

I sucked at combinations and permutations problems so am i screwed

is it similar flavour?

i have not learned any combinatorics beyond very basics

[delicate orchid]

basically, if you're given a representation and post-compose it with the trace map, this gives you a map G -> C, this map is the character of the represntation

these turn out to completely classify the representations up to isomorphism

[tardy hedge]

that sounds cool

[delicate orchid]

and they form a ring that's isomorphic to the ring of representations (including behaving very nicely with induction, restriction, and lifting)

with an inner product that tells you how representations decompose that you can compute incredibly easily

so instead of worrying about module theory you can just add little tuples together

[tardy hedge]

that all sounds cool. Im excited to learn more algebra

[coral spindle]

They're both extensions of C_2 by C_4 with the same action 🗿

Trivial by clifford theory

[delicate orchid]

method of little groups............

[coral spindle]

nerd_emoji

[delicate orchid]

wait

how are they the same action? D_8 splits, Q_8 doesn't

,w 2nd degree cohomology of C_4

[coral spindle]

The action of C_2 on C_4 is the same: inversion

[cloud walrus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by wewladstbh

[delicate orchid]

useless

[delicate orchid]

yeah there's not really any other choice

I just don't like it

[coral spindle]

So the Clifford theory is the same (barring some questions about extension of characters) since the action only matters in the quotient

It's so sick

[delicate orchid]

what the hell is a clifford...

[coral spindle]

The trace of a matrix is independent of the basis. So suppose you have a composition series for a finite-dimensional module of a k-algebra. Choose an appropriate basis, and you get block upper triangular matrices for every element of the algebra, and in particular taking the traces of these matrices will give you something depending only on the composition series.

In the rep theory of groups we look at the group algebra kG which is a k-algebra whose modules are representations of the group. If we have a module, we can take the trace of the action of each element of G. This gives us a function (typically not a homomorphism) chi : G → k, which is known as a character.

Fact: let k be a field of characteristic 0. Then two kG-modules are isomorphic if and only if their characters are equal.

Fact: Let k = C. There is a certain Hermitian inner product on 'class functions' which are functions G → k which are equal on conjugate elements. Then the set of characters of the simple modules form an orthonormal basis for the space of class functions.

These are two amazing results in elementary character theory.

[delicate orchid]

I thought you were typing up a response to this. I am disappointed

[coral spindle]

A clifford is a big red dog

OK now for the clifford stuff

I can't neglect my poor wew

[delicate orchid]

yeah just a slight elaboration would be nice

[coral spindle]

The two important results for me in clifford theory are a theorem in Isaacs which iirc isn't named, and Gallagher's theorem

So first

[delicate orchid]

I know Gallaghers

[tardy hedge]

that does sound epic. I dont know modules yet but im taking a course in them this fall

[delicate orchid]

figured out the difference though, one is a central extenstion

[coral spindle]

Let $N \unlhd G$. Note that $G$ acts on $\operatorname{Irr}(N)$ by conjugation. Given $\chi \in \operatorname{Irr}(N)$, let $G_\chi = \build{g \in G}{{}^g\chi = \chi}$ -- Isaacs calls this the inertia group.

There is a bijection $\operatorname{Ind}{G\chi}^G \colon \operatorname{Irr}(G_\chi \mid \chi) \to \operatorname{Irr}(G \mid \chi)$.

sigh

[cloud walrus]

Boytjie

[delicate orchid]

inertia group is standard terminology

[coral spindle]

I kinda hate it ngl

[delicate orchid]

and did you know that bijection holds block-wise?

[coral spindle]

I didn't! But I was aware that there is some nice Clifford theory of blocks

[delicate orchid]

yurrr

[delicate orchid]

I actually figured out the difference this time

[coral spindle]

I'm proud of u Wew

[delicate orchid]

they're not extenstions of C_4. They're extenstions of C_2^2

D_8 is the split, Q_8 isn't

[coral spindle]

Wait I'm confused

There's a sequence 1 → C_4 → G → C_2 → 1 in both cases

[delicate orchid]

D_8 can be written as C_2 \wr C_2

[coral spindle]

Ah yes I see

[delicate orchid]

or C_2^2 \ltimes C_2

[coral spindle]

So this is a twist you've done

[delicate orchid]

swaggg